Reply
Tue 30 Apr, 2019 08:53 am

I've been teaching myself Probability Mathematics, but I'm still struggling.

Please help me understand with an example.

Say I roll five six-sided dice all at once, with die faces numbered 1 through 6.

I want to determine the probability of AT LEAST three 6s occurring. First, is this the best way to set up the equation?

P(at least three 6s) = 1 - P(zero 6s) - P(exactly one 6) - P(exactly two 6s)

Or can the equation be better setup by subtracting AT LEAST occurrences?

I know this much...

P(zero 6s) = 5/6 * 5/6 * 5/6 * 5/6 * 5/6 = 3,125/7,776 = 0.40188

P(all 6s) = 1/6 * 1/6 * 1/6 *1/6 * 1/6 = 1/7,776

P(at least one 6) = 1 - P(zero 6s) = 4,651/7,776

To restate my whole problem, I'm unclear on the best way to calculate for "exactly 2" occurrences. Likewise for "exactly 3" or any other "exact" occurrence more than one but less than equal to the total number of dice thrown.

Thanks for your help!

@billy23,

Exactly 2 sixes is 6C2*(1/6)^2*(5/6)^4

Exactly 3 sixes is 6C3*(1/6)^3*(5/6)^3

I hope this helps.

@Fruityloop,

Quote:Exactly 2 sixes is 6C2*(1/6)^2*(5/6)^4

Exactly 3 sixes is 6C3*(1/6)^3*(5/6)^3

I hope this helps.

With 5 dice

Exactly 2 sixes is 5C2*(1/6)^2*(5/6)^3

Exactly 3 sixes is 5C3*(1/6)^3*(5/6)^2

https://en.wikipedia.org/wiki/Binomial_distribution
@knaivete,

Exactly 2 sixes is 5C2*(1/6)^2*(5/6)^3

Exactly 3 sixes is 5C3*(1/6)^3*(5/6)^2

I hope this helps.

5 dice not 6 dice.